The solubility of AgCl in water at 25^0 C is found to be 1.06*10^-5 moles per litre. Calculate the solubility product of AgCl at this temperature.
The solubility of AgCl in water at 250 C is found to be 1.06*10-5 moles per litre. Calculate the solubility product of AgCl at this temperature.
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AgCl ionizes completely in the solution as AgCl ----> Ag+ +Cl—
One mole of AgCl in the solution gives 1 mole of Ag+ ions and 1 mole of Cl—ions.
Now as the solubility of AgCl= 1.06*10—5 moles per litre.
Therefore [Ag+]=1.06*10—5 moles per litre and [Cl--] = 1.06*10—5 moles per litre.
Ksp for AgCl=[Ag+] [Cl--]=1.06*10—5 *1.06*10—5 =1.1*10—10
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